(Solved) equation of the tangent line t to the graph of F at the given point.

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Sample Q

Find the equation of the tangent line T to the graph of f at the given point. Use this linear approximation to complete the table. (Round your answers to four decimal places.)

f (x) = 2 / x² , (2, 1/2)

T (x) = Neutral Face

Solution

Derivative:
f '(x) = ( x^2 * 0 - 2 * 2x ) / (x^2)^2
f '(x) = ( 0 - 4x ) / x^4
f '(x) = ( - 4x ) / x^4
f '(x) = - 4/x^3 ----> x = 2
f '(2) = - 4/2^3
f '(2) = - 4/8
f '(2) = - 1/2

f(x) = 2/x^2 ----> x = 2
f(2) = 2/2^2
f(2) = 2/4
f(2) = 1/2

L(x) = T(x) = f(2) + f ' (2) (x - 2)
L(x) = T(x) = (1/2) - (1/2) (x - 2)

f(x) = 2/x^2 -----> x = 1.9
f(1.9) = 2 / (1.9)^2
f(1.9) ≈ 0.5540

L(x) = T(x) = (1/2) - (1/2) (x - 2) ---> x = 1.9
L(1.9) = T(1.9) = (1/2) - (1/2) (1.9 - 2)
L(1.9) = T(1.9) = 0.5500

f(x) = 2/x^2 -----> x = 1.99
f(1.99) = 2 / (1.99)^2
f(1.99) ≈ 0.5050

L(x) = T(x) = (1/2) - (1/2) (x - 2) ---> x = 1.99
L(1.99) = T(1.99) = (1/2) - (1/2) (1.99 - 2)
L(1.99) = T(1.99) = 0.5050

f(x) = 2/x^2 -----> x = 2
f(2) = 2 / (2)^2
f(2) = 0.5000

L(x) = T(x) = (1/2) - (1/2) (x - 2) ---> x = 2
L(2) = T(2) = (1/2) - (1/2) (2 - 2)
L(2) = T(2) = 0.5000

f(x) = 2/x^2 -----> x = 2.01
f(2.01) = 2 / (2.01)^2
f(2.01) ≈ 0.4950

L(x) = T(x) = (1/2) - (1/2) (x - 2) ---> x = 2.01
L(2.01) = T(2.01) = (1/2) - (1/2) (2.01 - 2)
L(2.01) = T(2.01) = 0.4950

f(x) = 2/x^2 -----> x = 2.01
f(2.1) = 2 / (2.1)^2
f(2.1) ≈ 0.4535

L(x) = T(x) = (1/2) - (1/2) (x - 2) ---> x = 2.1
L(2.1) = T(2.1) = (1/2) - (1/2) (2.1 - 2)
L(2.01) = T(2.01) = 0.4500

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equation of the tangent line t to the graph of F at the given point.

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